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Chapter 6: Problem 16

Determine which pairs of vectors in Exercises \(15-18\) are orthogonal. $$ \mathbf{u}=\left[\begin{array}{r}{12} \\ {3} \\ {-5}\end{array}\right], \mathbf{v}=\left[\begin{array}{r}{2} \\ {-3} \\ {3}\end{array}\right] $$

Short Answer

Expert verified
The vectors \( \mathbf{u} \) and \( \mathbf{v} \) are orthogonal.

Step by step solution

01

Understand the Definition of Orthogonal Vectors

Two vectors are orthogonal if their dot product is zero. The dot product can be calculated using the formula: \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \).
02

Identify Components of Both Vectors

The vectors \( \mathbf{u} \) and \( \mathbf{v} \) have components: \( \mathbf{u} = [12, 3, -5] \) and \( \mathbf{v} = [2, -3, 3] \).
03

Calculate the Dot Product

Substitute the components into the dot product formula: \( \mathbf{u} \cdot \mathbf{v} = (12 \times 2) + (3 \times -3) + (-5 \times 3) \).
04

Perform the Multiplication

Calculate each term of the product: \( 12 \times 2 = 24 \), \( 3 \times -3 = -9 \), and \( -5 \times 3 = -15 \).
05

Sum the Product Terms

Add the results from Step 4: \( 24 + (-9) + (-15) \).
06

Determine if Vectors Are Orthogonal

Calculate the sum: \( 24 - 9 - 15 = 0 \). Since the dot product is zero, the vectors \( \mathbf{u} \) and \( \mathbf{v} \) are orthogonal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product
The dot product, also known as the scalar product, is a fundamental operation in linear algebra. It's a tool used to measure the similarity and angle between two vectors. To compute the dot product of two vectors, each component of the first vector is multiplied by the corresponding component of the second vector. Then, all these products are summed up to reach a single scalar value. For vectors \( \mathbf{u} = [u_1, u_2, u_3] \) and \( \mathbf{v} = [v_1, v_2, v_3] \), the formula is: \[ \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \]
  • Multiplication of corresponding components
  • Summation of the results
If the dot product results in zero, the vectors are orthogonal, which means they meet at a right angle.
Linear Algebra
Linear algebra is a branch of mathematics that focuses on vector spaces and linear mappings between these spaces. It provides the tools and framework necessary for the representation of geometrical concepts using algebra. At its core, linear algebra is about vectors and the operations that can be performed on them. Concepts like orthogonality, which is the condition of being perpendicular, arise naturally in this discipline. Some key areas where linear algebra is utilized include:
  • Computer graphics for transformations and rendering
  • Machine learning for algorithms involving high-dimensional data
  • Physics, in modeling systems and solving equations
Vector Operations
Vectors have numerous operations that can be performed on them, each offering valuable insights into the properties of the vectors. One such operation, as already noted, is the dot product, which helps to determine if two vectors are orthogonal. Other important vector operations include:
  • Vector Addition: Combine two vectors by adding their corresponding components.
  • Scalar Multiplication: Multiply a vector by a scalar, scaling its magnitude without changing its direction.
  • Cross Product: For 3D vectors, results in another vector that is perpendicular to the original two.
These operations allow us to manipulate and analyze vectors in many dimensions, providing tools necessary for complex calculations and problem-solving in mathematics and physics.

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Most popular questions from this chapter

[M] To measure the takeoff performance of an airplane, the horizontal position of the plane was measured every second, from \(t=0\) to \(t=12 .\) The positions (in feet) were: \(0,8.8,\) \(29.9,62.0,104.7,159.1,222.0,294.5,380.4,471.1,571.7\) \(686.8,\) and \(809.2 .\) a. Find the least-squares cubic curve \(y=\beta_{0}+\beta_{1} t+\) \(\beta_{2} t^{2}+\beta_{3} t^{3}\) for these data. b. Use the result of part (a) to estimate the velocity of the plane when \(t=4.5\) seconds.

Let \(\mathbf{u}_{1}=\left[\begin{array}{r}{1} \\ {1} \\\ {-2}\end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{r}{5} \\ {-1} \\\ {2}\end{array}\right],\) and \(\mathbf{u}_{3}=\left[\begin{array}{l}{0} \\ {0} \\\ {1}\end{array}\right] .\) Note that \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) are orthogonal but that \(\mathbf{u}_{3}\) is not orthogonal to \(\mathbf{u}_{1}\) or \(\mathbf{u}_{2} .\) It can be shown that \(\mathbf{u}_{3}\) is not in the subspace \(W\) spanned by \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2} .\) Use this fact to construct a nonzero vector \(\mathbf{v}\) in \(\mathbb{R}^{3}\) that is orthogonal to \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2} .\)

Let \(\mathbf{u}_{1}, \ldots, \mathbf{u}_{p}\) be an orthogonal basis for a subspace \(W\) of \(\mathbb{R}^{n},\) and let \(T : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\) be defined by \(T(\mathbf{x})=\operatorname{proj}_{W} \mathbf{x}\) Show that \(T\) is a linear transformation.

Suppose 5 out of 25 data points in a weighted least-squares problem have a \(y\) -measurement that is less reliable than the others, and they are to be weighted half as much as the other 20 points. One method is to weight the 20 points by a factor of 1 and the other 5 by a factor of \(\frac{1}{2} \cdot\) A second method is to weight the 20 points by a factor of 2 and the other 5 by a factor of \(1 .\) Do the two methods produce different results? Explain.

Let \(\mathbf{u}=\left[\begin{array}{l}{a} \\ {b}\end{array}\right]\) and \(\mathbf{v}=\left[\begin{array}{l}{1} \\ {1}\end{array}\right] .\) Use the Cauchy-Schwarz inequality to show that $$\left(\frac{a+b}{2}\right)^{2} \leq \frac{a^{2}+b^{2}}{2}$$
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